[LeetCode]127 单词接龙II

题目描述

给定两个单词（beginWord 和 endWord）和一个字典，，找出所有从 beginWord 到 endWord 的最短转换序列。转换需遵循如下规则：

  1. 每次转换只能改变一个字母。
  2. 转换过程中的中间单词必须是字典中的单词。

说明:

  1. 如果不存在这样的转换序列，返回 0。
  2. 所有单词具有相同的长度。
  3. 所有单词只由小写字母组成。
  4. 字典中不存在重复的单词。
  5. 你可以假设 beginWord 和 endWord 是非空的，且二者不相同。

示例1：

输入:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]
输出:
[
  ["hit","hot","dot","dog","cog"],
  ["hit","hot","lot","log","cog"]
]

示例2：

输入:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]
输出: []
解释: endWord "cog" 不在字典中，所以不存在符合要求的转换序列。

思路

核心思想和LeetCode126一样，只是保存每一条路径，最后输入二维数组

代码

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class Solution:
    def findLadders(self, beginWord, endWord, wordList):
        """
        :type beginWord: str
        :type endWord: str
        :type wordList: List[str]
        :rtype: List[List[str]]
        """
        def bfs(front_level, end_level, is_forward, word_set, path_dic):
            if len(front_level) == 0:
                return False
            if len(front_level) > len(end_level):
                return bfs(end_level, front_level, not is_forward, word_set, path_dic)
            for word in (front_level | end_level):
                word_set.discard(word)
            next_level = set()
            done = False
            while front_level:
                word = front_level.pop()
                for c in 'abcdefghijklmnopqrstuvwxyz':
                    for i in range(len(word)):
                        new_word = word[:i] + c + word[i + 1:]
                        if new_word in end_level:
                            done = True
                            add_path(word, new_word, is_forward, path_dic)
                        else:
                            if new_word in word_set:
                                next_level.add(new_word)
                                add_path(word, new_word, is_forward, path_dic)
            return done or bfs(next_level, end_level, is_forward, word_set, path_dic)

        def add_path(word, new_word, is_forward, path_dic):
            if is_forward:
                path_dic[word] = path_dic.get(word, []) + [new_word]
            else:
                path_dic[new_word] = path_dic.get(new_word, []) + [word]

        def construct_path(word, end_word, path_dic, path, paths):
            if word == end_word:
                paths.append(path)
                return
            if word in path_dic:
                for item in path_dic[word]:
                    construct_path(item, end_word, path_dic, path + [item], paths)


        front_level, end_level = {beginWord}, {endWord}
        path_dic = {}
        wordSet = set(wordList)
        if endWord not in wordSet:
            return []
        bfs(front_level, end_level, True, wordSet, path_dic)
        path, paths = [beginWord], []
        construct_path(beginWord, endWord, path_dic, path, paths)
        return paths